Trigonometric Formulae, Equations and Inequalities (Senior 5 Mathematics): free lessons | Ganzaa

Trigonometric Formulae, Equations and Inequalities

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Senior 5 · Mathematics

Trigonometric Formulae, Equations and Inequalities

5 lessons

Lesson 1 of 5

Periods and the addition formulae

25 min

Is $\sin(30^\circ + 45^\circ)$ equal to $\sin 30^\circ + \sin 45^\circ$ ?

A calculator says $\sin 75^\circ \approx 0.9659$ . But $\sin 30^\circ + \sin 45^\circ = 0.5 + 0.7071 \approx 1.2071$ , which is bigger than 1, and no sine is bigger than 1. So the sine of a sum is not the sum of the sines. In S4 (Fundamentals of Trigonometry) you met the six ratios, the unit circle, the special angles and the graphs of $\sin x$ , $\cos x$ and $\tan x$ . This unit builds the formulae that connect the ratios of different angles, and uses them to solve equations and inequalities. It starts with the one property every trigonometric graph shares: it repeats.

Definition

Periodic function: a function $f$ for which there is a number $T \gt 0$ such that

f(x + T) = f(x) \quad \text{for every } x \text{ in the domain of } f.

Period: the smallest such positive number $T$ . One piece of the graph of length $T$ , repeated again and again, gives the whole graph.

Example: $\sin(x + 2\pi) = \sin x$ and no smaller positive number works, so $\sin x$ has period $2\pi$ . In the same way $\cos x$ has period $2\pi$ , while $\tan x$ and $\cot x$ have period $π \pi$ .

The period of $A\sin(ax + b) + c$

Take $a \ne 0$ . Adding $\frac{2\pi}{|a|}$ to $x$ adds $\pm 2\pi$ to the angle $a x + b$ , and the sine of the angle does not change:

\sin\left(a\left(x + \frac{2\pi}{|a|}\right) + b\right) = \sin(ax + b \pm 2\pi) = \sin(ax + b).

No smaller positive shift works, because the angle must move by a whole multiple of $2\pi$ . The numbers $A$ (which stretches the graph upwards) and $c$ (which lifts it) do not change the period. So

Function ( $a \ne 0$ )	Period
$A\sin(ax + b) + c$ and $A\cos(ax + b) + c$	$\frac{2\pi}{\|a\|}$
$A\tan(ax + b) + c$ and $A\cot(ax + b) + c$	$\frac{\pi}{\|a\|}$

Figure 1. Doubling x squeezes the wave:

\sin 2x

repeats every

π \pi

, and

\sin x

every

2\pi

A sum of two periodic functions. If $f$ has period $T_{1}$ and $g$ has period $T_{2}$ , and some number $T$ is a whole multiple of both, then $f + g$ repeats after $T$ . The smallest common multiple of the two periods is usually the period of the sum; check it on the graph or by testing a few values.

Worked example 1: finding periods

Find the period of (a) $f(x) = 3\sin\left(2x - \frac{\pi}{3}\right)$ (b) $g(x) = 5\tan\left(\frac{x}{2} + 1\right)$ (c) $h(x) = \sin 2x + \cos 3x$ .

Solution

Here $a = 2$ , so the period is $\frac{2\pi}{2} = \pi$ . The 3 and the $-\frac{\pi}{3}$ do not matter.
Here $a = \frac{1}{2}$ and the function is a tangent, so the period is $\frac{\pi}{\frac{1}{2}} = 2\pi$ .
$\sin 2x$ has period $π \pi$ and $\cos 3x$ has period $\frac{2\pi}{3}$ . The multiples of $π \pi$ are $\pi, 2\pi, 3\pi, \dots$ and the multiples of $\frac{2\pi}{3}$ are $\frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi, \dots$ . The smallest number in both lists is $2\pi$ , so $h$ has period $2\pi$ .

Warning

A power can change the period. $\sin x$ has period $2\pi$ , but $\sin^2 x$ has period $π \pi$ , because $\sin^2(x + \pi) = (-\sin x)^2 = \sin^2 x$ . In the same way $|\cos x|$ has period $π \pi$ . Do not copy the period of $\sin x$ to $\sin^2 x$ without checking.

The subtraction formula for cosine

Put two points on the unit circle (radius 1, centre O): $P(\cos A, \sin A)$ at angle $A$ and $Q(\cos B, \sin B)$ at angle $B$ . By the distance formula, and since $\cos^2 A + \sin^2 A = 1$ ,

PQ^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 2 - 2(\cos A\cos B + \sin A\sin B).

Now turn the whole picture clockwise about O through the angle $B$ . The point Q lands on $Q^{'} (1, 0)$ and P lands on $P'\big(\cos(A - B), \sin(A - B)\big)$ . A turn does not change lengths, so

PQ^2 = P'Q'^2 = \big(\cos(A - B) - 1\big)^2 + \sin^2(A - B) = 2 - 2\cos(A - B).

Figure 2. Turning the picture does not change the length of the chord, so

P Q = P^{'} Q^{'}

The two expressions for $P Q^{2}$ are equal, which gives the first addition formula:

\cos(A - B) = \cos A\cos B + \sin A\sin B \quad \text{for all real } A, B.

Definition

Compound angle: an angle written as a sum or difference of two angles, such as $A + B$ or $A - B$ .

Addition (compound-angle) formulae: identities that give the sine, cosine or tangent of a compound angle in terms of the ratios of $A$ and $B$ .

Example: $75^\circ = 45^\circ + 30^\circ$ is a compound angle, and its sine can be found exactly from the ratios of $45^\circ$ and $30^\circ$ .

The other five formulae, one from another

$\cos(A + B)$ : replace $B$ by $- B$ . Since $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$ , we get $\cos(A + B) = \cos A\cos B - \sin A\sin B$ .
$\sin(A + B)$ : use $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ : $\sin(A + B) = \cos\left(\left(\tfrac{\pi}{2} - A\right) - B\right) = \cos\left(\tfrac{\pi}{2} - A\right)\cos B + \sin\left(\tfrac{\pi}{2} - A\right)\sin B = \sin A\cos B + \cos A\sin B.$
$\sin(A - B)$ : replace $B$ by $- B$ again: $\sin(A - B) = \sin A\cos B - \cos A\sin B$ .
$\tan(A \pm B)$ : divide $\sin(A \pm B)$ by $\cos(A \pm B)$ , then divide the top and the bottom by $\cos A\cos B$ .

Important

The addition and subtraction formulae (for all real $A$ and $B$ ; the tangent formula needs $\cos A$ , $\cos B$ and $\cos(A \pm B)$ to be non-zero)

\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B

\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B

\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Tip

Signs: sine keeps the sign ( $+$ on the left, $+$ on the right) and mixes the ratios (sin cos, cos sin). Cosine flips the sign ( $+$ on the left, $-$ on the right) and keeps like with like (cos cos, sin sin).

Worked example 2: exact values

Without a calculator, find (a) $\sin 75^\circ$ (b) $\tan 15^\circ$ (c) $\cos\frac{7\pi}{12}$ .

Solution

$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$ , which matches the calculator.
$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \approx 0.2679$ .
$\frac{7\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}$ , so $\cos\frac{7\pi}{12} = \frac{\sqrt{2}}{2}\cdot\frac{1}{2} - \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4} \approx -0.2588$ . It is negative, as it should be for an angle between $\frac{\pi}{2}$ and $π \pi$ .

Worked example 3: angles in different quadrants

$\sin A = \frac{4}{5}$ with $\frac{\pi}{2} \lt A \lt \pi$ , and $\cos B = \frac{12}{13}$ with $\frac{3\pi}{2} \lt B \lt 2\pi$ . Find $\sin(A + B)$ , $\cos(A + B)$ and $\tan(A + B)$ .

Solution First the missing ratios, with signs from the quadrants. A is in the second quadrant, so $\cos A = -\sqrt{1 - \frac{16}{25}} = -\frac{3}{5}$ . B is in the fourth quadrant, so $\sin B = -\sqrt{1 - \frac{144}{169}} = -\frac{5}{13}$ .

\sin(A + B) = \frac{4}{5}\cdot\frac{12}{13} + \left(-\frac{3}{5}\right)\left(-\frac{5}{13}\right) = \frac{48 + 15}{65} = \frac{63}{65}

\cos(A + B) = \left(-\frac{3}{5}\right)\frac{12}{13} - \frac{4}{5}\left(-\frac{5}{13}\right) = \frac{-36 + 20}{65} = -\frac{16}{65}

\tan(A + B) = \frac{63}{65} \div \left(-\frac{16}{65}\right) = -\frac{63}{16}

Check with the tangent formula: $\tan A = -\frac{4}{3}$ , $\tan B = -\frac{5}{12}$ , and $\frac{-\frac{4}{3} - \frac{5}{12}}{1 - \frac{20}{36}} = \frac{-\frac{21}{12}}{\frac{4}{9}} = -\frac{63}{16}$ . ✓ Also $\left(\frac{63}{65}\right)^2 + \left(\frac{16}{65}\right)^2 = \frac{3969 + 256}{4225} = 1$ . ✓

Worked example 4: proving an identity

Prove that $\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B$ .

Solution Expand both factors; the product has the form $(p + q) (p - q) = p^{2} - q^{2}$ :

\sin(A + B)\sin(A - B) = (\sin A\cos B)^2 - (\cos A\sin B)^2 = \sin^2 A\,(1 - \sin^2 B) - (1 - \sin^2 A)\sin^2 B

= \sin^2 A - \sin^2 A\sin^2 B - \sin^2 B + \sin^2 A\sin^2 B = \sin^2 A - \sin^2 B. \quad \blacksquare

Warning

Never split a ratio over a sum. $\sin(A + B) \ne \sin A + \sin B$ and $\tan(A + B) \ne \tan A + \tan B$ . One check is enough to see it: with $A = B = \frac{\pi}{4}$ , $\sin\frac{\pi}{2} = 1$ but $\sin\frac{\pi}{4} + \sin\frac{\pi}{4} = \sqrt{2}$ .

Khan Academy: Proof of angle addition formula for sine (about 8 min). Sal proves the formula for sin(x + y) with right-angled triangles, a different route from the unit-circle proof above. Watch for how one length is split into two pieces, each giving one term of the formula.

✦

Practice

Work without a calculator unless told otherwise. The last two are harder.

Find the period of (a) $4\cos(3x + 1)$ (b) $\tan 4x$ (c) $-2\sin\frac{x}{3}$ (d) $\sin x + \sin\frac{x}{2}$ .
Find the exact value of $\cos 105^\circ$ .
Find the exact value of $\tan\frac{5\pi}{12}$ .
Simplify $\cos\left(\frac{\pi}{3} + x\right) + \cos\left(\frac{\pi}{3} - x\right)$ .
$A$ and $B$ are acute, with $\tan A = 2$ and $\tan B = 3$ . Find $\tan(A + B)$ , and deduce the angle $A + B$ .
$\sin A = \frac{5}{13}$ with $A$ acute, and $\cos B = -\frac{3}{5}$ with $B$ obtuse. Find $\sin(A - B)$ and $\cos(A + B)$ .
Prove that $\cos(A + B)\cos(A - B) = \cos^2 A - \sin^2 B$ .
Show that $\sin^2 x$ has period $π \pi$ and not a smaller one. (Hint: where is $\sin^2 x = 0$ ?)

Show answersHide answers

(a) $\frac{2\pi}{3}$ (b) $\frac{\pi}{4}$ (c) $6\pi$ (d) periods $2\pi$ and $4\pi$ , so $4\pi$ .
$\cos(60^\circ + 45^\circ) = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4} \approx -0.2588$ .
$\tan\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$ .
The $\sin$ terms cancel: $2\cos\frac{\pi}{3}\cos x = \cos x$ .
$\tan(A + B) = \frac{2 + 3}{1 - 6} = -1$ . Since $0 \lt A + B \lt \pi$ and the tangent is $- 1$ , $A + B = \frac{3\pi}{4}$ .
$\cos A = \frac{12}{13}$ , $\sin B = \frac{4}{5}$ . $\sin(A - B) = \frac{5}{13}\left(-\frac{3}{5}\right) - \frac{12}{13}\cdot\frac{4}{5} = -\frac{63}{65}$ ; $\cos(A + B) = \frac{12}{13}\left(-\frac{3}{5}\right) - \frac{5}{13}\cdot\frac{4}{5} = -\frac{56}{65}$ .
The product is $\cos^2 A\cos^2 B - \sin^2 A\sin^2 B = \cos^2 A(1 - \sin^2 B) - (1 - \cos^2 A)\sin^2 B = \cos^2 A - \sin^2 B$ .
$\sin^2(x + \pi) = (-\sin x)^2 = \sin^2 x$ , so $π \pi$ is a period. If $0 \lt T \lt \pi$ were a period, then $\sin^2 T = \sin^2 0 = 0$ , so $T$ would be a multiple of $π \pi$ , which is impossible. So the period is $π \pi$ .

Sources and credits

Written by Ganzaa: the explanations, the derivation, the diagrams, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S5 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Proof of angle addition formula for sine, Khan Academy, embedded from YouTube.

1Is sin⁡(30∘+45∘) \sin(30^\circ + 45^\circ) sin(30∘+45∘) equal to sin⁡30∘+sin⁡45∘ \sin 30^\circ + \sin 45^\circ sin30∘+sin45∘?

2The period of Asin⁡(ax+b)+c A\sin(ax + b) + c Asin(ax+b)+c

3The subtraction formula for cosine

4The other five formulae, one from another

Is $\sin(30^\circ + 45^\circ)$ equal to $\sin 30^\circ + \sin 45^\circ$ ?

The period of $A\sin(ax + b) + c$

The subtraction formula for cosine

The other five formulae, one from another